# Why is the standard deviation of the sample mean less than the population SD?

The sample means do not vary as much as the individual values in the population. That the sample means are less variable than the individual values in the population follows directly from the fact that each sample mean averages together all the values in the sample. A population consists of individual outcomes that can take on a wide range of values, from extremely small to extremely large. However, if a sample contains an extreme value, although this value will have an effect on the sample mean, the effect is reduced because the value is averaged with all the other values in the sample. As the sample size increases, the effect of a single extreme value becomes smaller because it is averaged with more values.

(Excerpt from my stat book)

I find this contradictory because they said that a sample mean averages together all values in the sample, but a population mean also averages together all the values. Also, they said that an extreme value's effect will be reduced when the sample size increases as it is averaged with more values. But in that case, since the population has the widest range doesn't the same thing apply even more?

So, why is the sample mean's standard deviation less than that of the population?

• The excerpt compares the variability of sample means with the variability of individuals (which are also sample means, for samples of size 1). Nov 11, 2014 at 22:42
• To be more sensical, and useful for future readers, should the question be renamed and edited to be something like "why is the standard error of the sample mean less than the standard deviation of the population?" That seems to be the focus of the extract. Nov 12, 2014 at 10:25
• I think this is a good question (+1) in part because the quoted argument implies the sample mean from any distribution with undefined mean (such as the Cauchy) would still be less dispersed than random values from that distribution, which is not true. (In fact, the sample means can exhibit greater dispersion than the original population.) Thus there is something subtly wrong about the argument. In particular, we should mistrust the appeal to the vague notion of "averaging together" that underlies this passage. (In defense of the textbook, it may elsewhere have assumed the SD exists.)
– whuber
Nov 12, 2014 at 16:02
• @whuber, the sample mean of the Cauchy distribution is the same as the original Cauchy distribution. Is there proof/example that a second or higher moment of a sample mean (even if it is not finite) can be larger than the same order moment of the distribution from which it is sampled? Sep 12, 2017 at 1:07
• @Martijn Consider any Student t distribution with parameter between $0$ and $1$. Or consider any distribution whose survival function decreases more slowly than $1/x$, such as a log-Cauchy. These have substantial chances of exhibiting huge outliers. The mean of a sample therefore has a greater chance of such an outlier than any individual draw from the distribution will have. When the tails are very long, the expected size of an outlier is so great that it more than compensates for the division by the sample size in computing the mean.
– whuber
Sep 12, 2017 at 14:27

I find this contradictory because they said that a sample mean averages together all values in the sample, but a population mean also averages together all the values.

The excerpt never says anything about the population mean.

since the population has the widest range doesn't the same thing apply even more?

Absolutely. If you took the mean of the entire population then it would have even less variability. But that has nothing to do with what the excerpt is talking about.

So, why is the sample mean's standard deviation less than the population?

This is explained in the exerpt.

• It seems I haven't gotten my questions and concepts right. will think things through and post and edit/update/comment Nov 11, 2014 at 19:04

Because the mean is an arithmetic midpoint and so when you add the deviations from the mean the sum will always be zero - hence the need to square the deviations BUT the median is an ordered midpoint so the sum of the deviations will ONLY be zero in a perfectly normal distribution. Hence, when you square the deviations you will always get a bigger value for the sum of squared deviations from the median than the mean... the rest flows from there...

• What you say is true but it does not answer the question. It is about the sample standard deviation which for independent identically distributed observations will tend to 0 as n gets large. Sep 12, 2017 at 1:05

Sorry for not going into each of your question. Still I would suggest if you can understand the below concept it will be easy for you to get out of this confusion

1.Suppose the mean of any sample represents each number of the sample which is influenced largely by extreme numbers.

2.With this, observe each extreme number in any sample is now being represented by a mean of that sample which is not as extreme as the number itself. Most of the times they are suppressed in opposite direction.

3.If you do this for several samples coming out of same population, in general you will observe sample means have less variability than individual numbers because calculating mean is taming the numbers towards their sample mean and ultimately towards population mean.

This is the reason standard deviation of the sample means is less than the population SD.

• You probably didn't notice, but the comment thread after the question concerns circumstances where this argument is incorrect.
– whuber
Aug 7, 2018 at 11:50